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The size of a television screen is given as the length

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The size of a television screen is given as the length

by Gmat_mission » Mon Jun 18, 2018 1:58 am

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The size of a television screen is given as the length of the screen's diagonal. If the screens were flat, then the area of a square 21-inch screen would be how many square inches greater than the area of a square 19-inch screen?

A. 2
B. 4
C. 16
D. 38
E. 40

[spoiler]OA=E[/spoiler].

I am confused here. Could anyone give me some help, please? I don't understand how to solve this PS question.
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by Vincen » Mon Jun 18, 2018 2:38 am
Hello Gmat_mission.

Let's take a look at your question.

We know that the size of a television screen is given as the length of the screen's diagonal.

Now, if d is the diagonal of a square, then we get $$d^2=l^2+l^2=2l^2\ \ \Rightarrow\ \ l^2=\frac{d^2}{2}$$ where "l" is the length of the side of the square. Also, we know that the area of a square is equal to l^2, therefore we get: $$Area=\frac{d^2}{2}.$$ Now, the area of a square 21-inch screen is $$Area=\frac{21^2}{2}$$ and the area of a square 19-inch screen is $$Area=\frac{19^2}{2}.$$ To know how many inches greater is the first screen with respect to the second one we have to subtract both areas as follows: $$Area_1-Area_2=\frac{21^2}{2}-\frac{19^2}{2}=\frac{21^2-19^2}{2}=\frac{\left(21-19\right)\left(21+19\right)}{2}$$ $$=\frac{\left(2\right)\left(40\right)}{2}=40.$$ So, the correct answer is the option E.

I hope this explanation may help you.
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by GMATGuruNY » Mon Jun 18, 2018 2:45 am
Gmat_mission wrote:The size of a television screen is given as the length of the screen's diagonal. If the screens were flat, then the area of a square 21-inch screen would be how many square inches greater than the area of a square 19-inch screen?

A. 2
B. 4
C. 16
D. 38
E. 40
We can save a little time if we know the following:

The area of a rhombus with diagonals d� and d₂ = (d�d₂)/2.
A square is a rhombus with 4 equal angles.
In a square, the two diagonals are equal.
Thus, the area of a square with diagonal d = d²/2.

a² - b² = (a+b)(a-b).

Thus:
Big TV - Little TV = 21²/2 - 19²/2 = (1/2)(21² - 19²) = (1/2)(21+19)(21-19) = 40.

The correct answer is E.
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by Brent@GMATPrepNow » Mon Jun 18, 2018 4:30 am
Gmat_mission wrote:The size of a television screen is given as the length of the screen's diagonal. If the screens were flat, then the area of a square 21-inch screen would be how many square inches greater than the area of a square 19-inch screen?

A. 2
B. 4
C. 16
D. 38
E. 40
Let x be the length (and width) of the square screen with diagonal 21
The area of the large screen will be x²

Let y be the length (and width) of the square screen with diagonal 19
The area of the small screen will be y²

Our goal is to find the value of x² - y²

Large TV: If we examine the right triangle created by 2 sides (both with length x) and the diagonal, we can apply the Pythagorean Theorem to get x² + x² = 21²
When we simplify this, we get 2x² = 441, which means x² = 441/2

Small TV: If we examine the right triangle created by 2 sides (both with length y) and the diagonal, we can apply the Pythagorean Theorem to get y² + y² = 19²
When we simplify this, we get 2y² = 361,, which means y² = 361/2

We can now find the value of x² - y²
We get x² - y² = 441/2 - 361/2 = 80/2 = 40

Answer: E

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by Scott@TargetTestPrep » Wed Jun 20, 2018 4:04 pm
Gmat_mission wrote:The size of a television screen is given as the length of the screen's diagonal. If the screens were flat, then the area of a square 21-inch screen would be how many square inches greater than the area of a square 19-inch screen?

A. 2
B. 4
C. 16
D. 38
E. 40
Let's determine the side of the square 21-inch screen (i.e., the diagonal of the screen is 21 inches). Recall that the diagonal of a square is equal to side√2.

21 = side√2

21/√2 = side

Since area is side^2, the area of the 21-inch screen is (21/√2)^2 = 441/2.

Let's determine the side of the square 19-inch screen:

19 = side√2

19/√2 = side

The area of the 19-inch screen is (19/√2)^2 = 361/2.

Thus, the difference is 441/2 - 361/2 = 80/2 = 40.

Alternate solution:

We are given two square TV screens with diagonals 21 and 19, respectively. We have to determine the difference between the areas of the screens. Recall that the area of a square, given its diagonal d, is A = d^2/2. Thus, the area of the 21-inch screen is 21^2/2 = 441/2 and the area of the 19-inch screen is 19^2/2 = 361/2. Therefore, the difference in areas is 441/2 - 361/2 = 80/2 = 40.

Answer: E

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